Proof of Gauss's Law

leright

It seems you're having trouble grasping divergence.

The divergence of a point in space is the NET outward flux through an arbitrarily small volume, per unit volume, enclosing this point, where the small volume approaches zero. BY DEFINITION, a radial field HAS NO DIVERGENCE except at r = 0. All flux going into all arbitrarily small volumes must go out of the small volume, causing a net outward flux of zero, or, a divergence of zero at all points. However, if you determine the divergence at the point where there is charge, you are at the point where flux lines EMANATE, so through an arbitrarily small volume enclosing this point there IS a net outward flux. Gauss's law in differential form states that the divergence of the electric flux density is equal to the volume charge density at that point.

We can apply the integral form of Gauss's law and find the net outward flux through a surface enclosing a charge distribution, and then take the limit of the outward flux as the volume approaches zero. However, you might notice that this limit is zero. In fact, this limit is zero for all charge distributions and all fields, which is quite obvious. This is where divergence of a point comes in. It is, as I stated above, the net outward flux of an arbitrarily small volume, PER UNIT VOLUME, enclosing a point in a vector field. IF we find the limit of the integral form of gauss's law to a volume where the size of that volume goes to zero about a point, the result is zero. However, if we find the divergence, we get the limit of the charge per unit volume, at that point, which is a finite number.

The derivation for Gauss's law, in its most simple form, is a calculation of the net outward flux through a closed surface, where that closed surface encloses a point charge, and this calculation is performed using a closed surface with radial symmetry, and the center of this surface is centered at the point charge. Due to symmetry, E is a constant and da is a constant, and both vectors E and da are at an angle of 0 degrees, so the flux simply becomes Q/ep_0. So, we have proved that for a point charge with a spherical closed surface enclosing it, the net outward flux of the E-field is Q/ep_0. We can rewrite this by finding the net outward flux of the D-field for this configuration, and we simply get Q.

The issue, as you stated, is applying this to all charge distributions and all volumes, ans gauss's law says we can do this. Well, I suppose you could just say that no matter what surface you have enclosing the charge, your net ourward flux will be the same, and therefore enclosed charge will be the same. But this is where I bring in intuition....

Gauss's law is just something that is second nature to me I suppose.

Galileo

Not true. Radial means that [itex]\vec F(\vec r)=F(r)\hat r[/itex], so the field just points away radially from the origin (or wherever the source may be).
A field like [itex]\vec F(\vec r)=K \hat r/r^3[/itex] is radial, but not divergenceless. That is a property unique to the inverse square relationship. In fact, an inverse square field is the only radial field that is divergenceless.

leright

ok, this is true. There can be radial fields with divergence of zero everywhere, if for instance, the magnitude of the field at r = 0 is 0. I had my mind in electrostatics world and was visualizing the electric fields due to charges, and not just fields in general.

Swapnil

Thank you leright for your explanation. I think I have a more better understanding of divergence now.

I think I am wrong, but isn't [tex]\vec{r}[/tex] the follwoing in spherical coordinates?:

[tex] \vec{r} = r\,\hat{r} + \arctan(\frac{y}{x})\,\hat{\phi} + \arcsin(\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}})\,\hat{\theta}[/tex]

How is this more easier?

Galileo

No way. In spherical coordinates the field is simply [itex]\hat r/r^2[/itex]. There is no phi or theta dependence in there, since it's radial remember? All you have to do is lookup the divergence in spherical coordinates and do a simply derivative.

Swapnil

Oh yeah! Sorry, I guess I wasn't thinking straight now that I think about it...

jbusc

If you happen to have an E&M book handy, there's probably some derivation of the integral form of Gauss's law from the generalized coulomb's law, and then the derivation of the differential form from the integral form, which is fairly trivial using the divergence theorem.

There's a fairly rigorous derivation in Classical Electrodynamics, by Jackson p27-29, and a somewhat less-than-rigorous (but quite descriptive) derivation in Introduction to Electrodynamics, by Grffiths, p65-70

Meir Achuz

Try page 12 of "Classical Electromagnetism" by Franklin for a rigorous, intuitive, mathematical proof.

Swapnil

Checked out Jackson. I agree that he gives a fairly rigorous derivation of Gauss' law.

debkar

Hey i have a new proof of gauss law.now i am listing it below

let there be an arbitaraly shaped body.let there be a charge Q,e='epsilon not',E=electric field,dS=small surface element.
let the body be consisting of many small surface elements dS₁,dS₂......
let the individual surfaces be so small that radius here are constant and be r₁,r₂......... respectively
therefore surface integral of E.dS=E₁.dS₁+E₂.dS₂+.....
since theta =0 therefore E.dS=E*dS

therefore putting values

Q*dS₁/(4*pi*e*r₁²)+Q*dS₂/(4*pi*e*r₂²)+....

d(theta₁)=dS₁/dr₁²,d(theta₂)=dS₂/dr₂²,etc (from definition of solid angle)

taking everything in common


net elctric flux=Q/(4*pi*e)(dS₁/dr₁²+dS₂/dr₂²+...)
=Q/(4*pi*e)(d(theta₁)+d(theta₂)+....)
=Q/(4*pi*e)(4*pi)=Q/(e) (since sum of all solid angle around the charge is 4*pi)

GRDixon

In general the electric field of a point charge with arbitrary motion does NOT vary as 1/r^2. For a rigorous formula for the E field of a point charge, see Griffiths, "Introduction to Electrodynamics," 2nd Edition, Eq. 9.107. I have considered the case of a relativistically oscillating charge (wA=.99c) inside a variety of shapes ... sphere, ellipsoid, egg-shape ... and have found (using a computer) that Gauss' law was satisfied in each case. I must confess I was surprised! Others have argued that Gauss' law is one of Maxwell's equations, and is therefore completely general. It appears they're right. I'm convinced that the law holds for all charge motions, including relativistic ones, and for all (fixed) shapes in space (both those enclosing the charge and those external to the charge).

gabbagabbahey

Not sure why you're reviving a 4 year old thread, but I suppose if you are looking for the Electrodynamic (as opposed to electrostatic) analog to Griffiths' derivation, you could simply apply the Divergence theorem to Jefimenko's equation for [itex]\textbf{E}[/itex]. However, unlike the static case where both Gauss' Law and Coulomb's are easy enough to derive from experiments (empirically), and are hence both valid axioms for the theory, I've never seen Jefimenko's equations derived empirically. Hence, I'd consider Gauss' Law to be more fundamental than Jefimenko's equations (although I suppose you could transform Coulomb's law to a moving reference frame and take Coulomb's law and the two (or 3) axioms of SR as axioms for your theory of electrodynamics), and question the importance of "deriving" Gauss' Law from Jefimenko's equations.

experimentX

[itex] \vec E = \frac{\hat r}{4 \pi r^2} \times \frac{Q}{\epsilon_0} \\
\vec E \cdot \hat r 4 \pi r^2 = \frac{Q}{\epsilon_0} \\
\text{Total Flux} = \frac{Q}{\epsilon_0} \\
\oint_s \vec E \cdot \hat n ds = \frac Q{\epsilon_0}[/itex]
Which is integral form of Gauss law.

Killingtensor

I'm fairly sure this can be done with lagrangians as long as you are comfortable with the jump from lagrangians to lagrangian densities.